Calcrostic Solving Tips

Detective Work on Calcrostic Problems

Solving a calcrostic is like checking out a crime scene for clues. The following are examples, how to get clues from single lines of the puzzle.

For example, if the row, column or diagonal is

a × b = a

then
a
must be 0 or
b
must be 1. It is easy to find out which case applies. We look at other lines that include
a
and
b
. For example, if
a=0
then
c + a = c
and if
b=1
then
c + b = d
. Similarly, from
a ÷ b = a
follows
a=0
or
b=1
and from each of
cd × b = cd
,
cd ÷ b = cd
follows
b=1
.
a + b = cd

then it follows that
c=1
because the sum of two 1-digit numbers can not be more than 9+9=18 and if both are different then not more than 9+8=17. The same conclusion can be made from
cd − b = a
.
a × b = c

then what is special is that the result is only a one digit number, so less than 10. Also,
a, b, c
are all different, so none of them can be 1 or 0. Thus, one of
a, b
must be 2 and the other one 3 or 4 and
c
is 6 or 8. The same conclusion can be drawn from
c ÷ b = a
.
a × a = b

then
b
is a square number unequal
a
and
b<10
, so
a=2, b=4
or
a=3, b=9
.
c + ea = eg

then the first digit (the tens) in
ea
and in
eg
are the same, so we conclude that
c + a = g
.
c + ea = fg

then the first digit (the tens) in
ea
and in
fg
are different. This can only be due to a carry over, so we conclude that
e + 1 = f
and
c + a = 10 + g
.
If a number has more than one digit then the leftmost digit can be assumed to be non-zero. If a puzzle is large and has 10 different letters then this easy to get information may even be sufficient to tell which letter has the value 0.
..a × ..b = ..5

then
a
or
b
is 5 and the other one is an odd digit.
..a × ..b = ..7

then the only possible values of
a
and
b
are 3 and 9. The product of 1 and 7 does also end at 7 but if
a
or
b
would be 7 then this would be known.
..a × ..b = ..3

then the only possible values of
a
and
b
are 7 and 9. The product of 1 and 3 does also end at 3 but if
a
or
b
would be 3 then this would be known.
..a × ..a = ..9

then the only possible values of
a
are 3 and 7.
..a × ..b = ..1

then the only possible values of
a
and
b
are 3 and 7.
..a × ..a = ..1

then the only possible values of
a
are 1 and 9.
..a × ..b = ..a

This clue is the same as the very first clue but more general with more digits that may appear to the left of
a
and
b
. This clue appears relatively often. By looking only at the units position it follows that
a × b = a + k × 10
where
k
is the carry-over from the multiplication. It follows that
a × (b-1) = k × 10
. In other words,
a × (b-1)
must be divisible by 10! Apart from the two cases known from the first clue: (
a = 0
) or (
b-1 = 0
) we only have 2 more cases to consider: (
a = 5
and
b
is even) or (
b-1 = 5
and
a
is even).
If logical reasoning does not help to determine more values then one has to guess and consider different cases. If one only wants to find one solution and not all then one should consider most likely cases first. What is likely and what not? From the above discussion we learned that it is unlikely for the unit value of a product to be 7, 3 or 1.
All statements made above about products equally apply to quotients.
Left most digits of multiple digit numbers can not be zero.
If the calcrostic involves not only integers but also rational numbers then one can draw more conclusions:

The leftmost digit in numerator and denominator can not be zero. If the numerator or denominator is a single digit then this can not be zero either.
If the denominator is a single digit, then this is not equal 1.
Because numerator and denominator are co-prime, the unit digits of numerator and denominator can not both be even. Also, if both unit digits are the same letter then they can not be 5 and if they are not the same letter then it can not be the case that one is 0 and the other is 5.

Try to find more clues, for example:

What can be concluded from
a × a = ba
? Which values can
a, b
only have?
What follows from
eb × c = cd
? Which value can
e
only have?
If you know that
g = 1
then what does
fg ÷ c = d
tell you about
f, c, d
?

Let us solve one puzzle:

   ab + c = dd
    ×   −    −
    e + f =  c
    =   =    =
   fb ÷ e = ab

The last clue above applies to the first column. We therefore have 4 cases:

b=0
(not possible otherwise the first row should give
ab + c = ..c
),
e=1
(not possible otherwise in the first column
ab × 1 = ab
)
e-1=5
, i.e.
e=6
and
b
even: 1. column: if
ab × 6
is a 2-digit number (
fb
) then
a=1
because already 20×6=120 results in a 3-digit number. With
a=1
from the first row then follows
d=2
because the ten unit can increase only by 1 when adding a 1-digit number.
b
has to be even in this 3rd case but
b<>0, b≠2
because
d=2, b≠6
because
e=6, b≠8
because from 1st column 18×6>100. Therefore
b=4
, from 1st row
c=8
and from 2nd row
f=2
which contradicts
e=2
. Therefore case 3) does not apply.
b=5
and
e-1
is even, i.e.
e
is odd: It follows that
e=3
because
e≠1
(case 2),
e≠5
(because
b=5
),
e<7
(because in 1st column 15×7>100). For
a
we have the constraints:
a≠2
(because otherwise from 1st row follows
d=a+1=3
but we already have
e=3
) and
a<>3
(because
e=3
),
a<4
(because from 1st column 45×3>100 and not a 2-digit number). Therefore
a=1
, from 1st row
d=2, c=dd - ab = 22-15=7
, from 2nd row
f=c-e=7-3=4
giving us the solution
```
   15 + 7 = 22
    ×   −    −
    3 + 4 =  7
    =   =    =
   45 ÷ 3 = 15
```

Have fun when trying our problem of the day!

Video solutions with more hints are available for the following calcrostics problems from Caribou contests: